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The diophantine equation Ax p + By q = Cz r Frits Beukers October 6, 2004 1 Introduction Let A,B,C Z be non-zero and p, q, r Z 2 . Consider the diophantine equa- tion Ax p + By q = Cz r , gcd(x, y, z)=1 in the unknown integers x, y, z. The gcd-condition is really there to avoid triv- ialities. For example, from a + b = c it would follow, after multiplication by a 21 b 14 c 6 , that (a 11 b 7 c 3 ) 2 +(a 7 b 5 c 2 ) 3 =(a 3 b 2 c) 7 thus providing us with infinitely many trivial solutions. There are three cases to be distinguished. 1. The hyperbolic case 1 p + 1 q + 1 r < 1. In this case the number of solutions is at most finite, as shown in [DG, Theorem 2]. 2. The euclidean case 1 p + 1 q + 1 r =1. A simple calculation shows that the set {p, q, r} equals one of {3, 3, 3}, {2, 4, 4}, {2, 3, 6}. In this case the solution of the equation comes down to the determination of rational points on twists of genus 1 curves over Q with j =0, 1728. 3. The spherical case 1 p + 1 q + 1 r > 1. A simple calculation shown that the set {p, q, r} equals one of the following: {2, 2,k} with k 2 or {2, 3,m} with m =3, 4, 5. In this case there are either no solutions or infinitely many. In the latter case the solutions are given by a finite set of polynomial parametrisations of the equation, see [Beu] 1
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Page 1: The diophantine equation Axp Byq Czr

The diophantine equation Axp + Byq = Czr

Frits Beukers

October 6, 2004

1 Introduction

Let A,B,C ∈ Z be non-zero and p, q, r ∈ Z≥2. Consider the diophantine equa-tion

Axp +Byq = Czr, gcd(x, y, z) = 1

in the unknown integers x, y, z. The gcd-condition is really there to avoid triv-ialities. For example, from a + b = c it would follow, after multiplication bya21b14c6, that

(a11b7c3)2 + (a7b5c2)3 = (a3b2c)7

thus providing us with infinitely many trivial solutions. There are three casesto be distinguished.

1. The hyperbolic case1p

+1q

+1r< 1.

In this case the number of solutions is at most finite, as shown in [DG,Theorem 2].

2. The euclidean case1p

+1q

+1r

= 1.

A simple calculation shows that the set p, q, r equals one of 3, 3, 3, 2, 4, 4, 2, 3, 6.In this case the solution of the equation comes down to the determinationof rational points on twists of genus 1 curves over Q with j = 0, 1728.

3. The spherical case1p

+1q

+1r> 1.

A simple calculation shown that the set p, q, r equals one of the following:2, 2, k with k ≥ 2 or 2, 3,m with m = 3, 4, 5. In this case there areeither no solutions or infinitely many. In the latter case the solutions aregiven by a finite set of polynomial parametrisations of the equation, see[Beu]

1

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F.Beukers, The diophantine equation Axp + Byq = Czr 2

A special case of interest is when A = B = C = 1. In many such cases thesolution set has been found. Here is a list of exponent triples of solved equa-tions together with the non-trivial solutions, i.e. xyz 6= 0. We start withthe hyperbolic cases. The first case n, n, n is of course Wiles’s proof of Fer-mat’s Last Theorem. As is well-known this proof is based on the proof ofthe Shimura-Taniyama-Weil conjecture for stable elliptic curves. Later Breuil,Conrad, Diamond and Taylor proved the full conjecture for any elliptic curvein [BCDT]. In the following list the cases with variable n are all solved usingWiles’s modular form approach, with possibly a few exceptions which are re-solved using Chabauty’s method. The isolated cases in this table are all solvedusing a Chabauty approach.

n, n, n and n ≥ 4. It has been shown by Wiles and Taylor [W],[TW]that there are no non-trivial solutions (formerly Fermat’s Last Theorem).

(4, n, 4) and n ≥ 3. Darmon [D] showed there are no non-trivial solutions.

n, n, 2 Darmon and Merel [DM] showed that there are no non-trivialsolutions when n is a prime ≥ 7, and Poonen showed this for n = 5, 6, 9.

n, n, 3 Darmon and Merel [DM] showed that there are no non-trivialsolutions when n is a prime ≥ 7, Lucas (19th century) showed this forn = 4 and Poonen for n = 5.

3, 3, n For 17 ≤ n ≤ 10000 it was shown by Kraus [Kr1] that there areno non-trivial solutions. For n = 4, 5 this was shown by Bruin [Br2,3].

(2, n, 4) This can be dealt with by application of results from [BS]. Thereare no non-trivial solutions.

(2, 4, n) Ellenberg [El] showed there are no non-trivial solutions when n ≥211.

(2n, 2n, 5) Bennett [Ben] showed there are no non-trivial solutions whenn ≥ 7. The case n = 3 was proved by N.Bruin [Br3], n = 2 is elementary[Ben] and n = 5 is follwos from Fermat’s last theorem.

2, 4, 5 Solved by N.Bruin [Br2], the non-trivial solutions are 25+72 = 34,35 + 114 = 1222.

2, 3, 8 Solved by N.Bruin [Br1,Br2], the non-trivial solutions are 18 +23 = 32, 438 + 962223 = 300429072, 338 + 15490342 = 156133.

2, 3, 7 Solved by Poonen, Schaefer, Stoll (to be published), the non-trivial solutions read 17+23 = 32, 27+173 = 712, 177+762713 = 210639282

and 92623 + 153122832 = 1137.

There are a two more solutions of hyperbolic type known: 1k+23 = 32, 73+132 =29. Presumably these solutions, and the ones listed above, are the only solutionsin the hyperbolic case. The following conjecture was put forward by Tijdemanand Zagier and is now also known as Beal’s conjecture:

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Conjecture 1.1 The diophantine equation

xp + yq = zr

in x, y, z ∈ Z with gcd(x, y, z) = 1, xyz 6= 0 and p, q, r ∈ Z≥3 has no solutions.

In the euclidean case it is well-known that the only non-trivial solutions arisefrom the equality 16+23 = 22, as the elliptic curves x3+y3 = 1, y2 = x4+1, y2 =x3 ± 1 contain only finitely many obvious rational points.In the spherical cases the solution set is infinite. In the case 2, 2, k this isan exercise in number theory. The case 2, 3, 3 was solved by L.J.Mordell,2, 3, 4 by D.Zagier and 2, 3, 5 by J.Edwards [Ed] in 2004. The families ofsolutions are listed in Appendix A (please read the explanation in the beginningof Appendix A).

2 A sample solution

To illustrate the phenomena we encounter when solving the generalized Fermatequation, we give a partial solution of x2 + y8 = z3. This equation lends itselfvery well to a stepwise descent method.First we solve x2 + u2 = z3. By factorisation on both sides over Z[i] we quicklysee that x+iu should be the cube of a gaussian integer, (a+bi)3. By comparisonof real and imaginary parts we get x = a3 − 3ab2, u = b(3a2 − b2). Note thata, b should be relatively prime in order to ensure gcd(x, u, z) = 1.Next we partly solve x2 + v4 = z3. This can be done by requiring that u,as found in the previous equation should be a square, e.g. v2 = b(3a2 − b2).The two factors on the right should be squares up to some factors ±3, sincetheir product is a square and a, b are relatively prime. We should explore allpossibilities, but in this partial solution we only continue with the possibilityb = −v2

1 , 3a2−b2 = −v22 . The latter equation can be rewritten as 3a2 = b2−v2

2 .The right hand side factors as (b− v2)(b+ v2) and hence each factor is a squareup to a finite number of factors. Here several possibilities present themselvesagain and we choose one, namely b − v2 = −6a2

1, b + v2 = −2a22 (and of

course a = 2a1a2). Summation of the two equalities and use of b = −v21 gives

us v21 − a2

2 = 3a21. Now the left hand side factors and we choose the possibility

v1−a2 = 6t2, v1+a2 = 2s2 (and of course a1 = 2st). Solving for v1 and a2 givesv1 = s2 + 3t2 and a2 = s2 − 3t2. Hence a = 4st(s2 − 3t2) and b = −(s2 + 3t2)2.Further straightforward computation gives us

v = (s2 + 3t2)(s4 − 18s2t2 + 9t4)x = 4st(s2 − 3t2)(3s4 + 2s2t2 + 3t4)(s4 + 6s2t2 + 81t4)z = (s4 − 2s2t2 + 9t4)(s4 + 30s2t2 + 9t4)

A might be clear now, this gives us an infinite set of integer solutions to theequation x2 + v4 = z3. Had we followed all possibilities we would have foundmore parametrised solutions to recover the full solution set in integers. For a

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full list see Appendix A, or Henri Cohen’s recent book [Co], wher one finds acomplete derivation of the above type.Finally we solve x2 + y8 = z3. To that end we must solve

y2 = (s2 + 3t2)(s4 − 18s2t2 + 9t4).

After division by t6 and putting ξ = s/t, η = y/t3 we get

η2 = (ξ2 + 3)(ξ4 − 18ξ2 + 9),

i.e. we must determine the rational points on a genus two curve. To solve theequation completely we must determine the rational points on several genus twocurves, namely those arising from the different parametrising solutions above.To cut things short now, we can easily calculate that

z3

y8=

(ξ4 − 2ξ2 + 9)3(ξ4 + 30ξ2 + 9)3

η8.

Thus, any point z3/y8 coming from a solution of x2 + y8 = z3 is the image ofa rational point (ξ, η) on our genus two curve under the map just given. Thismap is an example of a Galois cover map.Had we followed all possibilities of the above argument, we would have obtaineda number of covering maps from a genus 2 curve to P1 which would have coveredthe full set of values z3/y8 corresponding to all solutions of x2 + y8 = z3 in co-prime integers x, y, z.In this example the curves arose naturally as a result of a descent procedure.In many cases, like x3 + y5 = z7 this descent is not so obvious any more andwe have to start by constructing covers of P1 by curves which have a suitableramification behaviour.

3 Galois covers of P1

In all approaches to the solution of the (generalised) Fermat equations one usesGalois covers in one form or another.First we recall a few facts from the theory of algebraic curves and their functionfields. For a more complete introduction we recommend Chapter II of Silver-man’s book [Si]. Let K be a field of characteristic zero and X a complete,smooth and geometrically irreducible curve X defined over K. In the functionfield K(X) we consider a non-constant element which we denote by φ. Notethat K(X) is now a finite extension of the field K(φ). The degree of this ex-tension is also called the degree of the map φ. Let P ∈ X(K) (by X(L) wedenote the L-rational points of X, where L is a field extension of K). Assumingfor the moment φ(P ) 6= ∞ we call the vanishing order of φ − φ(P ) at P theramification index of φ at P . Notation: eP . In case φ(P ) = ∞ we take for eP

the vanishing order of 1/φ at P . If eP > 1 we call P a ramification point of φ.The image φ(P ) under φ of a ramification point P is called branch point. Theset of branch points is called the branch set or branch locus. We now recall theRiemann-Hurwitz formula

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Theorem 3.1 With the notation above let N be the degree of the map φ andg(X) the geometric genus of X. Then,

2g(X)− 2 = −2N +∑

P∈X(K)

(eP − 1).

As we have eP = 1 for all points of X except finitely many, the sum on the rightis in fact a finite sum.We call the map given by φ a geometric Galois cover if the extensionK(X)/K(φ)is a Galois extension of fields. The Galois group G is a subgroup of the auto-morphism group (over K) of X and is called the covering group. Note that theextension K(X)/K(φ) need not be Galois. If it is we call the cover simply aGalois cover. For a geometric Galois cover the ramification indices of all pointsabove a given branch point are the same. In particular we shall be interestedin geometric Galois covers whose branch locus is 0, 1,∞. These are examplesof so-called Belyi maps. An immediate consequence of the Riemann-Hurwitztheorem is the following.

Corollary 3.2 . Let X → P1 be a geometric Galois cover whose branch locusis contained in the set 0, 1,∞. Suppose that above these points the ramificationindices are p, q, r. Suppose the degree of the cover is N . Then

2g(X)− 2 = N

(1− 1

p− 1q− 1r

).

In particular we see that if 1/p + 1/q + 1/r > 1, then g(X) = 0 and when1/p+ 1/q + 1/r < 1 we have g(X) ≥ 2.Here we list a series of geometric Galois covers that will occur in the sequel. Westart with X = P1. The finite subgroups of AutQ(P1) have been classified byFelix Klein. Up to conjugation they are given by

1. The cyclic group of order N

2. The dihedral group of order 2N

3. The tetrahedral group of order 12

4. The octahedral group of order 24

5. The icosahedral group of order 60

When we consider P1 as a sphere, each of these examples correspond to the finiterotation groups of the sphere. Here we describe them in some more detail, wherez denotes a standard coordinate on P1. We cannot go into all the fascinatingdetails of the Klein groups. For an extensive discussion we recommend ChapterI of Klein’s original book [Kl].Cyclic group. This group is generated by z 7→ ζNz where ζN is a primitiveN -th root of unity. The corresponding cover is given by z 7→ zN .

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Dihedral group. This is generated by the cyclic group given above and z 7→1/z. The cover is given by

z 7→ 12

(zN +

1zN

).

Tetrahedral group. Let ω be a primitive cube root of unity. Consider thesubgroup Γ3 of SL(2,C) generated by

1√−3

(1 2ωj

ω−j −1

)(j = 0, 1, 2) and

(ω 00 ω−1

).

Then the tetrahedral group is the subgroup of PSL(2,C) given by Γ3/±1. Thecovering map is given by

z 7→(

4(z3 − 1)z4 + 8z

)3

.

F.Klein’s (semi)-invariants of Γ3 are

f = −4y(x3 − y3)

H = −x4 − 8xy3

t = −x6 + 20x3y3 + 8y6

with fundamental relation t2 +H3 = f3.Octahedral group. Consider the group Γ4 generated by(

ζ8 00 ζ−1

8

),

(0 1−1 0

),

1√2

(ζ8 −ζ−1

8

ζ8 ζ−18

).

Then the octahedral group is the subgroup of PSL(2,C) given by Γ4/± 1. Thecover is given by

z 7→ (z8 + 14z4 + 1)3

108(z(z4 − 1))4.

F.Klein’s (semi)-invariants are

f = 36xy(x4 − y4)

H = −36(x8 + y8 + 14x4y4)

t = 216(x12 + y12 − 33(x4y8 + x8y4)

with fundamental relation t2 +H3 = −3f4.Icosahedral group. Consider the group Γ5 generated by

−Id,(ζ5 00 ζ−1

5

),

1√5

(ζ5 − ζ4

5 −ζ25 + ζ3

5

−ζ25 + ζ3

5 −ζ5 + ζ45

).

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Then the icosahedral group is the subgroup of PSL(2,C) given by Γ5/±1. Thecover is given by

z 7→ (−z20 + 228z15 − 494z10 − 228z5 − 1)3

1728z5(z10 + 11z5 − 1)5.

F.Klein’s (semi)-invariants are

f = 123xy(x10 + 11x5y5 − y10)

H = 124(−x20 − y20 + 228(x15y5 − x5y10)− 494x10y10)

t = 126(x30 + y30 + 522(x25y5 − x5y25)− 10005x20y10 − x10y20))

with the fundamental relation t2 +H3 = f5.In the last three examples the forms f,H, t have the additional property that

H =1

k2(k − 1)2

∣∣∣∣ fxx fxy

fxy fyy

∣∣∣∣ , t =1

2k(k − 2)

∣∣∣∣ fx fy

Hx Hy

∣∣∣∣ ,where k is the degree of f . These relations will become important later on.Furthermore in all three examples the branch locus is given by the points0, 1,∞ ∈ P1. The ramification indices above these points are 3, r, 2 wherer = 3, 4 or 5 depending on the group Γr

Now we turn to the case when the genus of X is at least 2 and list a number ofexamples.

1. X : xn + yn = zn and covering map (x : y : z) 7→ (x/z)n. This map hasdegree n2 and the group is given by al elements (x : y : z) 7→ (ζx : ζ ′y : z)where ζ, ζ ′ are n-th roots of unity. The branch locus is given by 0, 1,∞with ramification indices n, n, n.

2. Let p and q be integers ≥ 3 and let X be given by the projective equations

p−1∑i=0

ζikp x

qi = 0 (k = 1, 2, . . . , p− 2).

Consider the covering map

(x0 : x1 : . . . : xp−1) 7→(∑p−1

i=0 xqi )

p∏p−1i=0 x

qi

.

This has Galois group of order pqp−1 generated by multiplication of thecoordinates xi by a q-th root of unity and the cyclic permutation of thecoordinates (x0 : x1 : . . . : xp−1) 7→ (x1 : x2 : . . . : xp−1 : x0). Notice alsothat for points on X we have the relation

(p−1∑i=0

xqi )

p + (p−1∑i=0

ζ−ip xq

i )p = (

p−1∏i=0

xi)q.

The map has branch locus 0, 1,∞ and ramification indices p, p, q.

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3. Let n ≥ 2 and letX be the complete modular curveX(n). We consider thenatural map X(n) → X(1) = P1 using the J-function on X(n). More ex-plicitly, Consider the modular J-function on the complex upper half planeH. This map gives us the quotient map J : H → C with respect to thegroup PSL(2,Z). It ramifies above the points J = 0, 1 with ramificationindices 3 and 2 respectively. Let

Γ(n) = M ∈ SL(2,Z) | M ≡ Id (mod n).

Then Γ(n) is a normal subgroup of SL(2,Z) and the quotient of H by Γ(n)is denoted by Y (n). Since Γ(n) contains no elliptic elements, the coverH → Y (n) is unramified. Furthermore J factors over Y (n) to a finite mapJ : Y (n) → C. If we now complete the curves by adding the cusps to Y (n)and ∞ to C, we get J : X(n) → P1 where X(n) is the completion of Y (n).This map ramifies of order n above ∞. So the ramification indices above0, 1,∞ are 3, 2, n. The covering group is PSL(2,Z/nZ). When n = 3, 4, 5we recover the tetrahedral, octahedral and icosahedral covering again.

4. Let n be odd, X = X(2n) and consider the natural map to X(2) = P1.This has ramification indices n, n, n above 0, 1,∞ and no others. Thecovering group is PSL(2,Z/nZ).

5. Let n be odd and let X be the completed modular curve corresponding tothe modular group Γ(n)∩ Γ0(2). Then the natural map X → X0(2) = P1

is a geometric Galois cover ramified above 0, 1,∞ with ramification indicesn, n, 2. The covering group is again PSL(2,Z/nZ)

6. Similarly, when n is not divisible by 3 we consider the modular groupΓ(n)∩Γ0(3) and take for X the associated complete modular curve. ThenX → X0(3) gives us a geometric Galois cover ramified above 0, 1,∞ withramification indices n, n, 3. The covering group is PSL(2,Z/nZ).

4 Lifting points

Let φ : X → P1 be a geometric Galois cover defined over a number field K andwhose degree is N . For any point a ∈ P1(K) the points in the inverse imageφ−1(a) generate a finite Galois extension L of K of degree at most N . In thefollowing we explicitly determine the set of primes of K that ramify in L.Let π be any finite prime of K. We extend it to a valuation of K. We representpoints of P1(K) as points in K ∪∞. We define the π-adic intersection numberon P1 by

Iπ(a, b) =

ordπ(a− b) if ordπ(a), ordπ(b) ≥ 0ordπ(1/a− 1/b) if ordπ(1/a), ordπ(1/b) ≥ 0

0 otherwise

The following theorem is a weak version of a theorem proved in [Bec].

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Theorem 4.1 (S.Beckmann) Let X → P1 be a Galois cover defined over anumber field K and with covering group G. Let a1, . . . , ar ∈ K ∪∞ be the setof branch points. There is finite set of primes, which we denote by Sbad, withthe following properties. For any point q ∈ K not equal to any ai we have

1. the finite primes of K that ramify in K(φ−1(q)) are contained in the setS = Sbad ∪ Sq, where Sq is the set of primes π at which q meets a branchpoint ai π-adically.

2. if π 6∈ Sbad and q meets the branchpoint ai π-adically, then π ramifiesof order e where e is the denominator of Iπ(q, ai)/ei and where ei is theramification index above ai.

In [Bec] we find a stronger statement which explicitly gives us Sbad. If the groupG is simple or if the covering is given by a good model, then Sbad is the unionof the primes dividing the order of G and the primes for which distinct branchpoints have positive intersection.By a good model we mean the following. Let O be the ring of integers of K.Consider the extension K(X)/K(x) as before. Let O[X] be the integral closureof O[x] in K(X). We say that we have a good model if every prime π of O, forwhich the prime ideal π[x] ramifies in O[X], belongs to Sbad.We are now able to give a proof of the following result.

Theorem 4.2 Let φ : X → P1 be a geometric Galois cover which ramifiesof order p, q, r above the points 0, 1,∞ respectively, and which has no furtherramification. Suppose that the cover is defined over the number field K. Thenthere exists a finite extension L of K such that φ−1(Aap/Ccr) ⊂ X(L) for everytriple (a, b, c) that satisfies

Aap +Bbq = Ccr, gcd(a, b, c) = 1.

Here X(L) denotes the set of L-rational points on X.

Proof. If necessary we replace K by a finite extension so that φ becomes aGalois cover. Consider the field M generated over K by the coordinates of thepoints in φ−1(Aap/Ccr). We now apply Beckmann’s Theorem. We let SABC bethe set of primes dividing ABC. Let π be a prime of K not dividing abc and notin Sbad ∪ SABC . Then the point Aap/Ccr doesn’t reduce to 0, 1 or ∞ moduloπ. To see that it does not reduce to 1 notice that Aap

Ccr − 1 = −Bbq

Ccr . Hence π isunramified in M/K. Suppose now that π 6∈ Sbad∪SABC and π divides a. Thenthe intersection number Iπ(Aap/Ccr, 0) is a positive multiple of p. This is aconsequence of the fact that gcd(a, b, c) = 1. Since the cover ramifies of order pabove zero, part 2 of Beckmann’s theorem implies that π has ramification order1, i.e. no ramification. Similarly, if π divides b or c and is not in Sbad ∪ SABC ,then π is unramified in M/K. So we find that the coordinates of a point inφ−1(Aap/Ccr) are in a number field of degree at most N , the degree of thecover, and a fixed set of ramified primes. There are only finitely many suchfields and for L we can take their compositum.

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qed

We can now prove Theorem 2 in [DG]

Theorem 4.3 (Darmon-Granville) Suppose 1/p+1/q+1/r < 1 and A,B,C ∈Z with ABC 6= 0. Then the number of solutions to

Axp +Byq = Czr, gcd(x, y, z) = 1

is finite.

Proof. We begin by the construction of a curve X and a geometric Galoiscover X → P1 of Belyi-type, i.e it ramifies only above the points 0, 1,∞. Apossible way of doing this is via the Riemann existence theorem. Accordingto a theorem of A.Weil this cover can then be defined over a number field K.By the Riemann-Hurwitz theorem we know that 1/p + 1/q + 1/r < 1 impliesg(X) ≥ 2. Beckmann’s theorem implies that there is a number field L such thatfor any solution (a, b, c) we have φ−1(Aap/Ccr) ⊂ X(L). By Faltings’ theorem(formerly Mordell’s conjecture) we know that X(L) is finite. Hence our equationhas finitely many solutions.

qed

5 Galois cocycles

Let K be a number field and L a finite Galois extension. Let G be a finitegroup with a Gal(L/K) Galois action Gal(L/K) → Aut(G). A 1-cocycle is amap ξ : Gal(L/K) → G, mapping σ 7→ ξσ, such that

ξστ = ξσσ(ξτ )

for all σ, τ ∈ Gal(L/K). Two cocycles ξ, ζ are called cohomologous if there existsh ∈ G such that

ζσ = h−1ξσσ(h).

The set of cocycles modulo this equivalence relation is called the first Galoiscohomology set of Gal(L/K) in G. Notation H1(Gal(L/K), G).An important use of the first cohomology is the description of twists of algebraicvarieties V , when G = Aut(V ). To fix ideas, let X be a smooth connectedalgebraic curve defined over K. Any curve X ′ defined over K together with anisomorphism ψ : X → X ′, which is defined over K, is called a twist of X. Inparticular, when the twist map ψ is defined over a finite galois extension L ofK, we call our twist an L-twist. Let ψ : X → X ′ be such an L-twist. Then,for any σ ∈ Gal(L/K) the composite map ψ−1σ(ψ) is an automorphism of Xdefined over L. One easily checks that

σ 7→ ψ−1σ(ψ)

is a Galois cocycle in H1(Gal(L/K),AutL(X)). Namely,

ψ−1στ(ψ) = ψ−1σ(ψ)σ(ψ−1τ(ψ)).

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Two L-twists ψ1 : X → X ′ and ψ2 : X → X ′′ are called equivalent is thereexist h ∈ AutL(X) and an isomorphism g : X ′ → X ′′ defined over K such thatψ2 = g ψ1 h. Denote the set of classes of L-twists by Twist(X,L/K). Thenwe have

Theorem 5.1 The map ψ 7→ (σ 7→ ψ−1σ(ψ)) gives a well-defined map fromTwist(X,L/K) to H1(Gal(L/K),AutL(X)). Moreover, this map is a bijection.

More explicitly, if we have a 1-cocycle ξ : Gal(L/K) → AutL(X), then it ispossible to find an L-twist ψ : X → X ′ such that ξσ = ψ−1σ(ξ) for all σ ∈Gal(L/K). We now apply this to our diophantine equation.

Theorem 5.2 Let A,B,C, p, q, r be as in the introduction. By Sol we denotethe set of numbers Aap/Ccr for all a, c belonging to triples of integers (a, b, c)that satisfy

Aap +Bbq = Ccr, gcd(a, b, c) = 1, abc 6= 0

Let φ : X → P1 be a geometric Galois cover of Belyi-type which ramifies above0, 1,∞ of order p, q, r respectively. Suppose it is defined over a number field K.Then there exist finitely many twists ψi : X → Xi, i = 1, 2, . . . , r, defined overK, such that

1. each map φ ψ−1 : Xi → P1 is defined over K.

2. Sol ⊂ ∪ri=1φ ψ

−1i )(Xi(K)).

3. The sets φ ψ−1i (Xi(K)) intersect in a subset of 0, 1,∞.

Proof. According to Theorem 4.2 there is a finite Galois extension L such thatφ−1(Sol) ⊂ X(L). We assume that G is also defined over L. Take any pointQ ∈ Sol and let P ∈ X(L) be such that φ(P ) = Q. Since φ is a geometricGalois cover, for any σ ∈ Gal(L/K) there exists a unique gσ ∈ G such thatσ(P ) = gσ(P ). Notice that

gστ (P ) = σ(τ(P )) = σ(gτ (P )) = σ(gτ )(σ(P )) = σ(gτ )gσ(P ).

Hence gστ = σ(gτ )gσ and so we see that

σ 7→ g−1σ

is a Gal(L/K) cocycle in H1(Gal(L/K), G). Consider the twist ψ : X → X ′

that corresponds to this cocycle. This means that g−1σ = ψ−1σ(ψ) for all σ ∈

Gal(L/K). Hence

σ(ψ(P )) = σ(ψ)(σ(P )) = ψg−1σ gσ(P ) = ψ(P ).

In other words ψ(P ) is fixed under Gal(L/K) and hence ψ(P ) ∈ X ′(K). Fur-thermore, for any σ ∈ Gal(L/K) we have

σ(φ ψ−1) = φ σ(ψ)−1 = φ gσ ψ−1 = φ ψ−1.

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F.Beukers, The diophantine equation Axp + Byq = Czr 12

Hence φ ψ−1 is defined over K. Since Q = φ(P ), we see that Q is containedin φ ψ−1(X ′(K)).To every class inH1(Gal(L/K), G) we choose a twist and sinceH1(Gal(L/K), G)is finite, we get a finite number of twists ψi : X → Xi with i = 1, 2, . . . , r. Partone of our Theorem follows.To see the disjointness, suppose φ ψ−1

1 (X1(K)) and φ ψ−12 (X2(K)) have

a point Q ∈ P1(K), Q 6= 0, 1,∞ in common. For i = 1, 2 choose a pointPi ∈ Xi(K) such that Q = φ ψ−1

i (Pi). Then there exists k ∈ G such thatψ−1

1 (P1) = k ψ−12 (P2). Let ξi be the cocycle to which we associated ψi. Then

application of any σ ∈ Gal(L/K) yields

ξ−11,σ ψ

−11 (P1) = σ(k) ξ−1

2,σψ−12 (P2).

Replacing the right hand side,

ξ−11,σ ψ

−11 (P1) = σ(k) ξ−1

2,σ k−1ψ−11 (P1).

Since ψ−1(P1) has trivial stabilizer in G we conclude that

ξ1,σ = k−1 ξ2,σ σ(k)

for all σ ∈ Gal(L/K). Hence ξ1, ξ2 are cohomologous and the twists X1, X2 areequivalent.

qed

So to solve a generalised Fermat equation in the hyperbolic case it suffices todetermine the K-rational points on a finite set of curves of genus ≥ 2. It wouldbe nice if one could have K = Q. In fact this is how the equations with exponenttriples 2, 3, 7, 2, 3, 8 and A = B = C = 1 were solved in [PSS] and [B1],[B2].In the spherical cases p, q, r = 2, 3, 3, 2, 3, 4, 2, 3, 5 we have the Kleincovers of degree 12, 24, 60 respectively and X = P1. Hence the above theoremimplies that the solution set of a generalised Fermat equation in the sphericalcase is given by a finite (possibly empty) set of rational functions P1 → P1

defined over Q.In the following we shall carry out the program just sketched in detail for thespherical case (2, 3, 5).

6 Invariant theory of binary forms

Here we give a very quick introduction following Hilbert’s lectures from 1897.See [H]. In particular our approach will be very classical. The only differencebetween Hilbert’s and our representation is that we use k instead of n for thedegree of the base form.

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6.1 Definition and first examples

Let K be an algebraically closed field of characteristic zero. Consider a formf ∈ K[a,x] of the shape

f(a,x) =k∑

i=0

(k

i

)aix

k−i1 xi

2,

which we call the base form. We have two sets of polynomial variables, x =(x1, x2) and a = (a0, . . . , ak). For historical reasons the number k is called theorder of f . The group GL(2,K) acts on polynomials in x1, x2 as follows. Forany g ∈ GL(2,K) we replace the column vector x = (x1, x2)t by the componentsof the column vector g ·x. When we replace the variables x1, x2 in a polynomialh in this way, we denote the new polynomial by h g.Let C ∈ K[a,x]. We denote its dependence on a,x by writing it as C(f). Thepolynomial C(f) is called a covariant of f if there exists an integer p ≥ 0 suchthat

C(f g) = det(g)pC(f) gfor all g ∈ GL(2,K). We call p the weight of the covariant. A covariant whichdepends only on the aj is called an invariant. I.e I(a) ∈ K[a] is called aninvariant of weight p if

I(f g) = det(g)pI(f)

for all g ∈ GL(2,C).Since the action of g does not change degrees in the ai and xj we can restrictour attention to covariants which are homogeneous in the aj and homogeneousin the xi. When C(f) is such a bihomogeneous covariant, we call dega(C) thedegree of C and degx(C) the order of C. Notice that f itself is a covariant ofweight 0, order k and degree 1.Here are two of our most important examples of covariants. First there is theHessian covariant H(f) defined by

H(f) =1

k2(k − 1)2

∣∣∣∣ f11 f12f21 f22

∣∣∣∣where fij stands for partial differentiation with respect to xi and xj . It is amatter of straightforward calculus to see that this is a covariant. Its weight is2, the order is 2k − 4 and the degree is 2.The other important covariant is the Jacobian determinant t(f) defined by

t(f) =1

k − 2

∣∣∣∣ f1 f2H1 H2

∣∣∣∣ .Again it is straightforward to check that this is a covariant. Its weight is 3, itsorder 3k − 6 and degree 3.

Remark 6.2 Let C be a covariant of f . When we specialise the variablesa0, . . . , ak to values in some ring R and we do this both in f and C(f) wewill still call the specialisation of C(f) a covariant of the specialised f .

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6.3 Structure of covariants

Suppose we are given a bihomogeneous polynomial

C(f) =m∑

j=0

(m

i

)Cj(a)xm−j

1 xj2

We give necessary and sufficient condition for a form to be a covariant. Supposeit is a covariant. Since GL(2,K) is generated by the matrices(

λ 00 1

),

(0 11 0

),

(1 ν0 1

)where λ ∈ K∗, ν ∈ K, it suffices to verify the covariant property of C only forthese matrices. First we take g to be the diagonal matrix with entries λ, 1. Then

g(x1) = λx1, g(x2) = x2.

Let Aar00 · · · ark

k xm−j1 xj

2 be a non-trivial term in C. In shorthand notation:Aarxm−j

1 xj2. The covariant property now implies that

λkr0+(k−1)r1+···+rk−1arxm−j1 xj

2 = λp+m−jarxm−j1 xj

2.

Hencekr0 + (k − 1)r1 + · · ·+ rk−1 = p+m− j.

The covariant property with respect to(

0 11 0

)implies if Aar0

0 · · · ark

k xm−j1 xj

2

occurs as a non-trivial term, then so does (−1)pAar0k · · · ark

0 xm−j2 xj

1. In partic-ular, this observations together with previous one, leads to

r1 + 2r2 + · · · krk = p+ j

for any monomial. Addition of the two equalities gives us

k(r0 + r1 + · · ·+ rk) = 2p+m

Letting g be the degree (in a) of C we get

kg = 2p+m.

Finally we need to implement the covariant property with respect to(

1 ν0 1

).

It is a straightforward but slightly tedious job to show that we get

DC = x2∂C

∂x1

where D is the differential operator

D = a0∂

∂a1+ 2a1

∂a2+ 3a2

∂a3+ · · ·+ nan−1

∂an.

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By the symmetry(

0 11 0

)we also get

∆C = x1∂C

∂x2

where ∆ is the differential operator

∆ = an∂

∂an−1+ 2an−1

∂an−2+ · · ·+ na1

∂a0.

A particular consequence of the first equation is that

D(C0) = 0. (1)

The second equation implies that

C1 =1m

∆C0, C2 =1

m(m− 1)∆2C0, . . . Cm =

1m!

∆mC0. (2)

In fact, these conditions turn out to be both necessary and sufficient. In thefollowing statement an isobaric polynomial in the aj is a polynomial such thatfor all terms Aar0

0 · · · ark

k the sum r1 + 2r2 + 3r3 + · · ·+ krk has the same value.

Theorem 6.4 The bihomogeneous polynomial

C(f) =m∑

j=0

(m

i

)Cj(a)xm−j

1 xj2

is a covariant of weight p if and only if C0 is homogeneous of degree g, isobaricof weight p, such that m = kg − 2p, and such that equations (1) and (2) aresatisfied.

In particular we have a very nice corollary characterising invariants.

Corollary 6.5 A homogeneous polynomial C(a) is an invariant of weight p ifand only if it has degree g and is isobaric of weight p such that kg = 2p andsuch that the equation DC(a) = 0 is satisfied.

6.6 Further examples

First we give some examples of invariants and covariants for small k.

The case k = 2, f = a0x21 + 2a1x1x2 + a2x

22.

The Hessian of f equals a0a2 − a21, the discriminant of f . It turns out that all

invariants are powers of the discriminant.

The case k = 3, f = a0x31 + 3a1x

21x2 + 3a2x1x

22 + a3x

22.

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The Hessian now reads

H(f) = (a0a2 − a21)x

21 + (a0a3 − a1a2)x1x2 + (a1a3 − a2

2)x22.

There is also the Jacobian covariant

t(f) = (a20a3 − 3a0a1a2 + 2a3

1)x31 + · · ·

The discriminant of f is an invariant,

D(f) = a20a

23 − 3a2

1a22 + 4a3

1a3 + 4a0a32 − 6a0a1a2a3.

The powers of D form a full system of invariants. We have the classical relation

4H3 + t2 = Df2.

The case k = 4, f = a0x41 + 4a1x

31x2 + 6a2x

21x

22 + 4a3x1x

32 + a4x

42.

We have the Hessian and Jacobian covariants H(f), t(f) as before. The ring ofinvariants is generated by

I2 = a0a4 − 4a1a3 + 3a22

I3 = a0a2a4 − a0a23 − a2

1a4 + 2a1a2a3 − a32

We have the classical relation

t(f)2 = −4H(f)3 + I2H(f)f2 − I3f3.

A general way to produce new covariants from old ones is the transvectantconstruction. Letting C1, C2 be two covariants and r ∈ Z≥1 we define

(C1, C2)r =(

(k − r)!k!

)2

Ωr(C1(x1, x2)C2(x′1, x′2))|x′1=x1,x′2=x2

whereΩ =

∂x1

∂x′2− ∂

∂x′1

∂x2.

The transvectants of f are defined by

τ2m =12(f, f)2m, τ2m+1(f) = (f, τ2m(f))1.

This is the sequence of transvectants we find in [H, Ch I.8]. They are covariantsof degrees 2 and 3 respectively with weights equal to the index n in τn. Onenotes that H(f) = τ2(f), t(f) = τ3(f) and

H(f) = (a0a2 − a21)x

2k−41 + · · ·

t(f) = (a20a3 − 3a0a1a2 + 2a3

1)x3k−61 + · · ·

τ4(f) = (a0a4 − 4a1a3 + 3a22)x

2k−81 + · · ·

τ6(f) = (a0a6 − 6a1a5 + 15a2a4 − 10a23)x

2k−121 + · · ·

The following theorem will be crucial to us.

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Theorem 6.7 (Gordan, 1887) The fourth transvectant τ4(f) of a non-trivialform f with k ≥ 4 is identically zero if and only if f is GL(2,K)-equivalent toone of the following forms

1. xk1 or xk−1

1 x2 (degenerate case)

2. x2(x31 + x3

2) (tetrahedral case)

3. x1x2(x41 + x4

2) (octahedral case)

4. x1x2(x101 − 11x5

1x52 − x10

2 ) (icosahedral case)

So the vanishing of τ4(f) forces f to be one of the Klein forms if f is notdegenerate.Because of its importance we give a proof of this theorem. First of all a straight-forward computation shows that τ4(f) vanishes for all forms in the list. We nowshow the converse statement. Let f be a form with τ4(f) = 0. Very explicitlywe have

τ4(f) =2n−8∑r=0

Drx2n−r1 xr

2

where

Dr =∑

i+j=r

(n− 4i

)(n− 4j

)(aiaj+4 − 4ai+1aj+3 + 3ai+2aj+2).

We use the equations D0 = 0, D1 = 0, D2 = 0, . . . to recursively determine thecoefficients aj . Suppose our f is not equivalent to xk

1 . Then f should have azero of order ≤ k/2. By application of a GL(2,K) substitution, we can see toit that this zero becomes x2 = 0. In particular, a0 = 0.First suppose that a1 = 0. Choose t > 1 minimal so that at 6= 0. We have thatt ≤ k/2 because x2 = 0 is a zero of order ≤ k/2. Now note that for all t ≤ k−2,

D2t−4 = 3(k − 4t− 2

)2

a2t + · · ·

where the omitted terms all contain a factor ai with i < t. Since ai = 0 for alli < t it follows from D2t−4 = 0 that at = 0, a contradiction. So a1 cannot bezero.Now suppose, after normalisation if necessary, that a1 = 1. By application of ashift x1 → x1 + νx2, x2 → x2 we can see to it that a2 = 0. We now determinethe remaning ai recursively using the equations

D0 = a0a4 − 4a1a3 + a22 = 0

Dr = · · ·+ k

r

k − 4r − 1

(r − 4 +

12k

)a1ar+3 + · · · = 0 (r ≥ 1)

where the omitted terms all contain a0 or an ai with 2 ≤ i ≤ r+2. If the factorr − 4 + 12/k does not vanish for any r we get that a3 = a4 = . . . = ak = 0

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and we are in the case xk−11 x2. So we need that k divides 12 and 4 > 12/k.

Hence k = 4, 6 or 12. Take k = 12, the other cases being similar. We getthat a2 = a3 = . . . = a5 = 0 and choose a6 6= 0. By scaling we can see to itthat a6 = −11. Recursive solution of D4 = D5 = . . . = D9 = 0 shows thata7 = . . . = a10 = a12 = 0 and a11 = −1. Hence f = x1x2(x10

1 − 11x51x

52 − x10

2 ).

7 Mordell’s approach

As an example of the use of invariant theory in solving diophantine equationswe present Mordell’s method to solve the equation

x2 = −y3 +A2yz2 +A3z

3 (3)

in integers x, y, z with gcd(x, y, z) = 1. Mordell’s idea is to exploit the relation

t(f)2 = −4H(f)3 + I2H(f)f2 − I3f3

for quartic forms f . Given a solution x, y, z with z 6= 0 he constructs a quarticform f with invariants I2(f) = 4A2, I3(f) = 4A3 and such that f(1, 0) =z,H(1, 0) = y, t(1, 0) = 2x. When we write f in our standard form, this amountsto solving

(i) z = a0

(ii) y = a0a2 − a21

(iii) 2x = a20a3 − 3a0a1a2 + 2a3

1

(iv) 4A2 = a0a4 − 4a1a3 + 3a22

(v) 4A3 = a0a2a4 − a0a23 − a2

1a4 + 2a1a2a3 − a32

We start by setting a0 = z. From (3) it follows that x2 ≡ −y3(modz2). Hence−(xy−1)2 ≡ y(modz2). Now choose a1 integral so that a1 ≡ −xy−1(modz2).Then y + a2

1 is divisible by z = a0 and we can determine a2 from equation (ii).Rewrite the equation (iii) as

z2a3 = 2x+ 3a1y + a31.

To solve this, the right hand side should be divisible by z2. This is indeed thecase as follows from

2x+ 3a1y + a31 ≡ 2x+ 3(−xy−1)y + (−xy−1)3 ≡ −x(y3 + x2)y−3(modz2)

and from equation (3).We now determine a4 from equation (iv). With this value of a4, equation (v) isautomatically satisfied because of (3). We now see from equations (iv) and (v)that both a0a4 = za4 and (a0a2 − a2

1)a4 = ya4 are integer. Since z and y arerelatively prime this implies that a4 is an integer.Thus we know that to any solution of (3) we have a quartic form f with pre-scribed invariants 4A1, 4A2 such that f(1, 0) = z,H(1, 0) = y, t(1, 0) = 2x. Of

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course other specialisations of f,H, t will provide us with an infinity of solutionsto 3. Since the number of SL(2,Z)-classes of such forms is finite, we get a finitenumber of parametrising solutions of (3) that give the complete solution set.Notice that I2 is the fouth transvectant of f . If this vanishes and if I3 = 4 weget the identity t2 = −4H3 − 4f3. This is exactly the case for which Mordellprovides a full solution set in [Mo, Chapter 25].

8 Edwards’s approach

The main idea in Edwards’s paper [Ed] is to mimick Mordell’s technique to solvethe diophantine equation

x2 + y3 = dz5 (4)

in coprime integers x, y, z. Here d is a given non-zero integer. Let

f(x1, x2) = 123x1x2(x101 − 11x5

1x52 − x10

2 )

be the icosahedral form of F.Klein. Letting H and t be its Hessian and Jacobiancovariants, we get

(t/2)2 + H3 = f5. (5)

Definition 8.1 Let d be a non-zero integer. By C5(d) we denote the set ofGL(2,Q)-transforms of f which are of the form

f(x1, x2) =12∑

i=0

(12i

)aix

12−i1 xi

2,

such that

1. a0, . . . , a5, 7a6, a7, . . . , a12 ∈ Z for all i.

2.(t(f)/2)2 +H(f)3 = df5. (6)

where H(f) and t(f) are the Hessian and Jacobian covariants of f .

Notice that a6 is preceded by a 7 in this definition (and in all formulas to come).It turns out that the space of dodecahedral forms with a0, . . . , a5, 7a6, a7, . . . , a12 ∈Z is stable under SL(2,Z). From now on, when we speak of integer solutions,we will mean these variables to be integral.Because of the covariant property it follows from (5) that for any g ∈ GL(2,Q)we have for f := f g the identity

(t(f)/2)2 +H(f)3 = det(g)6f5.

So by taking det(g)6 = d we can see to it that we get parametrisations ofx2 + y3 = dz5.Our first goal is to prove the following theorem.

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Theorem 8.2 Let d be a non-zero integer. Let x, y, z ∈ Z be a coprime solutionof x2 + y3 = dz5. Then there exists a form f ∈ C5(d) such that

f(1, 0) = z, H(f)(1, 0) = y, t(f)(1, 0) = 2x. (7)

Proof . In what follows we shall write a form12∑

i=0

(12i

)aix

12−i1 xi

2

in the shape[a0, a1, . . . , a12].

When z = 0, we have x = ±1 and y = −1. We can immediately write down thecorresponding forms f . They read

[0,±1, 0, 0, 0, 0,−144d/7, 0, 0, 0, 0,∓(144d)2, 0].

So from now on we can assume z 6= 0. We first prove our theorem without therationality properties of the ai. Determine α, β ∈ Q such that f(α, β) = z/dand H(α, β) = y/d2. Determine γ, δ ∈ Q such that αδ − βγ = 1. Define

the dodecahedral form f by f = df g, where g =(α γβ δ

). Then, because

H(f) = H(df g) = d2H(f)g and t(f) = t(df g) = d3t(f)g we find that (6)is satisfied for our choice of f . Moreover, f(1, 0) = df(α, β) = z and similarlyH(f)(1, 0) = y. From x2 + y3 = dz5 and (6) it follows that t(f)(1, 0) = ±2x. Incase t(f)(1, 0) = −2x we take a new f equal to the old f(ix1, ix2). This doesnot change f,H but it does change t by a minus sign. We have found a solutionf for the equations (6) and (7). Notice that if f(x1, x2) is a solution, then so isf(x1 + λx2, x2) for any λ ∈ Q. So we still have some freedom in the choice off . Thus far everything has been done over Q. Our claim is that we can chooseλ in such a way that the coefficients ai satisfy the rationality and integralityproperties of the ai required by f being in C5(d).Equations (7) gives us the following equations in ai

z = a0

y = a0a2 − a21

2x = a20a3 − 3a0a1a2 + 2a3

1

precisely the same as in Mordell. We also need explicitly given necessary condi-tions on the ai for f to be equivalent to f . These are given by the vanishing of thefourth transvectant according to Gordan’s theorem. So we get the Di = 0 wherethe Di, i = 0, 1, . . . , 12 are the coefficients of τ4(f). In Appendix B, at the veryend we have reproduced the explicit equations Di = 0 for i = 0, . . . , 9. Thenwe must take a0 = z. For a1 we have complete freedom because of the freedomin λ above. We set a1 equal to a number in the residue class −xy−1(modz5).From H(1, 0) = y and t(1, 0) = 2x it follows that

a0a2 ≡ y + (xy−1)2 ≡ −dz5y−2 ≡ 0(modz5)a20a3 ≡ −x(x2 + y3)y−3 ≡ −dxz5y−3 ≡ 0(modz5)

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From this we observe that a2 and a3 are integers divisible by z4 and z3 respec-tively. We can now determine a4, a5, . . . recursively. Start with

0 = D0/1 = a0a4 − 4a1a3 + 3a22.

Hence a0a4 is an integer divisible by z3. Hence a4 is an integer divisble by z2.Similarly it follows from D1 = 0 that a5 is an integer divisible by z and fromD2 = 0 it follows that 7a6 ∈ Z. In

D3/56 = 0 = a0a7 − 6a2a5 + 5a3a4

a small miracle happens. There is no term a1a6 and we can now see that a0a7

is an integer divisble by z5. Hence a7 is divisible by z4. The equation

D4/14 = 0 = 5a0a8 + 12a1a7 − 6a2(7a6)− 20a3a5 + 45a24

poses a small problem because of the coefficient 5 in front of a0a8. However, byelimination of a6, a7 from D4 = D3 = D2 = 0 we obtain

a20a8 = 12a4a3a1 + 18a4a

22 − 24a2

3a2 + 4a5a3a0 − 9a24a0.

Now it follows that a8 is an integer divisible by z3. Continuing with D5 = D6 =D7 = 0 we find that a9, a10, a11 are integers as well. From D8 = D9 = 0 we seethat a0a12 and a1a12 are integers. Because a0, a1 are coprime, we conclude thata12 is integral.

qed

Up to a shift x1 → x1 + ax2, x2 → x2 the form f found in Theorem 8.2 isunique.

Theorem 8.3 Let d, x, y, z be as in Theorem 8.2. Let f1, f2 ∈ C5(d) be suchthat

f1(1, 0) = f2(1, 0) = z, H1(1, 0) = H2(1, 0) = y, t1(1, 0) = t2(1, 0) = 2x.

Then there exists an integer q such that f1(x1, x2) = f2(x1 + qx2, x2).

Proof. Notice that if f(x1, x2) has coefficients a0, a1, a2, . . ., then for any num-ber q the form f(x1 + qx2, x2) has coefficients a0, a1 + qa0, a2 + 2qa1 + q2a0, . . ..We distinguish two cases. First of all suppose that z = 0. Then, automatically,y = −1, x = ±1. From the proof of Theorem 8.2 it follows that a0 = 0, a1 =∓1. From D4 = a0a4 − 4a1a3 + 3a2

2 = 0 we see that a2 is even. Hence by asubstition of the form (x1, x2) → (x1 + qx2, x2) we can see to it that a2 = 0.The remaining ai are now uniquely determined from the equations Di = 0 andthe extra equation R1 = 0 (see Appendix B). This latter equation arises fromthe identity τ6(f) = 360df and it fixes the proper normalisation of a6.Now suppose that z 6= 0. We should have a0 = z. From D4 = a0a4 − 4a1a3 +3a2

2 = 0 it follows that a2 is even if a0 is even. We can now dedude fromthe equations H(1, 0) = y, t(1, 0) = 2x that a1 ≡ −xy−1(modz). So by a

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F.Beukers, The diophantine equation Axp + Byq = Czr 22

substitution (x1, x2) → (x1 + qx2, x2) we can see to it that 0 ≤ a1 < |z|. Thisdetermines a1 uniquely. The remaining ai are now determined uniquely as wellby the equations H(1, 0) = y, t(1, 0) = 2x and Di = 0.

qed

Corollary 8.4 Let d, x, y, z be as in Theorem 8.2. Suppose we have f1, f2 ∈C5(d) and integers a1, b1, a2, b2 such that

z = f1(a1, b1) = f2(a2, b2)y = H1(a1, b1) = H2(a2, b2)

2x = t1(a1, b1) = t2(a2, b2)

Then f1 and f2 are SL(2,Z)-equivalent. Moreover, if the last equation reads

t1(a1, b1) = 2x t2(a2, b2) = −2x

then f1 and f2 are GL(2,Z)-equivalent.

Proof. Choose c1, d1 ∈ Z such that a1d1 − b1c1 = 1 and put g1 =(a1 b1c1 d1

).

Then f1 g1 is a form in C5(d) which specialises together with its covariants atthe point (1, 0) to the solution x, y, z. We can choose g2 ∈ SL(2,Z) similarly.According to Theorem 8.2 the forms f1 g1 and f2 g2 are SL(2,Z) equivalent.This shows the first part of our Corollary.To show the second part, choose a g ∈ GL(2,Z) with determinant −1. Letf ′ = f g. Then H(f ′) = H(f) g and t(f ′) = −t(f) g because H has evenweight and t has odd weight. According to the first part of our Corollary, f ′2and f1 are SL(2,Z)-equivalent.

qed

We have now seen that all coprime solutions to x2 + y3 = dz5 arise fromparametrisations using forms from C5(d) and their covariants. It remains toshow that C5(d) consists of a finite number of SL(2,Z)-orbits and, if possible,compute these orbits.

9 Reduction of binary forms

Also in this section we follow the approach in [Ed], but with a few simplications.Consider a form f ∈ R[x1, x2] of degree k ≥ 3 in x1, x2. We assume once andfor all that it has distinct zeros. Choose a factorisation over C,

f =k∏

i=1

(νix1 − µix2).

There is some ambiguity in the normalisation of the linear factors for the mo-ment, but this will be cleared. For any t1, . . . , tk ∈ R>0 define φ = φ(f, t)

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F.Beukers, The diophantine equation Axp + Byq = Czr 23

by

φ(f, t) =k∑

i=1

t2i (νix1 − µix2)(νix1 − µix2).

This is a real quadratic form which is positive definite since its values for real(x1, x2) 6= (0, 0) are strictly positive. Strictly speaking φ also depends on theparticular factorisation of f we have chosen. Let us write φ(f, t) = Px2

1 −2Qx1x2 +Rx2

2 and let δ(f, t) = PR−Q2 be its determinant.

Lemma 9.1 For any g ∈ GL(2,R) we have

φ(f g, t) = φ(f, t) g and δ(f g, t) = det(g)2δ(f, t).

Proof. Note that the second is a consequence of the first, while the first isimmediate from the definitions.

qed

We define the Hermite determinant of f as

Θ(f) := mint:

∏i ti=1

δ(f, t)k/2.

Note that this minimum does not depend on the particular normalisation in thefactorisation in f . In [CS, Lemma 4.2] it is shown that the minimum is assumedat a uniquely determined point, which we denote by t0. The representativepoint of f is the point z0 ∈ H such that φ(f, t0)(z0, 1) = 0. Note also thatthis representative point is independent of the normalisation of the µi, νi. If therepresentative point of f is in the standard fundamental domain |z| ≥ 1,−1/2 ≤<(z) ≤ 1/2 we call f Hermite reduced.

Theorem 9.2 Let f be a real form of degree k ≥ 3 and distinct roots. Then,for any g ∈ GL(2,R) we have

1. Θ(f g) = det(g)kΘ(f).

2. If z0 is the representative point of f and z1 = g−1(z0) (fractional lineartransform) then the representative point of f g is given by z1 if det(g) > 0and z1 if det(g) < 0.

Proof. From δ(f g, t) = det(g)2δ(f, t) it follows that

Θ(f g) = min∏ti=1

δ(f g, t)k/2

= |det(g)|k min∏ti=1

δ(f, t)k/2

= |det(g)|kΘ(f)

Let t0 be the point t where the minimum is attained. Then from φ(f g, t0) =φ(f, t0) g it follows that

φ(f g, t0)(z1, 1) = (φ(f, t0) g)(z1, 1)= |γz1 + δ|2φ(f, t0)(z0, 1)

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F.Beukers, The diophantine equation Axp + Byq = Czr 24

where g =(α βγ δ

). Hence z1 is a zero of the quadratic form φ(f g, t0). When

det(g) > 0 this lies in the upper half plane, so it is the representing point off g. When det(g) < 0 however, the conjugate zero z1 lies in H.

qed

Theorem 9.3 Let f be a real from of degree k ≥ 3 and distinct roots withfactorisation f =

∏i(νix−µiy). Let z0 = x+ iy its representative point. Then,

Θ(f) =(k

2y

)k k∏i=1

(|νix− µi|2 + |νiy|2).

This Theorem allows us to compute the Hermite determinant of the form f(x1, x2) =123x1x2(x10

1 − x51x

52 − x10

2 ). Notice that f(x1, x2) = f(x2,−x1). Let z0 be therepresenting point of f(x1, x2). Then, by covariance, the representing pointof f(x2,−x1) is −1/z0. But by the invariance of the form f we should havez0 = −1/z0. Thus we conclude that z0 = i. Using our Theorem it is straight-forward to verify that Θ(f) = 22431855.

Theorem 9.4 Let f ∈ C5(d). Then

Θ(f) = 22431855|d|2.

Proof. There exists an element g ∈ GL(2,C) such that f = f g. In [Ed]it is shown that we can assume g ∈ GL(2,R). From the covariance of therepresenting point we have

Θ(f) = |det(g)|12Θ(f).

(Using the SL(2,C)-reduction theory developed in [CS] one deduces that this fol-lows also without the assumption g ∈ GL(2,R)). We also have that |det(g)|6 =d(f)/d(f) and we know that d(f) = 1. Hence we conclude

Θ(f) = |d|2Θ(f)

and our Theorem follows.qed

The next theorem gives us upper bounds for the coeffcients of Hermite reducedforms.

Theorem 9.5 Let

f =k∑

i=1

(k

i

)aix

k−i1 xi

2

be a real, Hermite reduced form of degree k. Then for all i+ j ≤ k we have

|aiaj | ≤(

43k2

)k/2

Θ(f).

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F.Beukers, The diophantine equation Axp + Byq = Czr 25

Proof. Let z0 ∈ H be the representing point of f and write z0 = x + iy. Lett1, . . . , tk be the components of the vector t that minimizes δ(f, t). We shallshow that for all r,

|ar|2 ≤|z0|2r

(ky)kΘ(f).

Recalling that y ≥√

32 max(|z0|, 1) when z0 is in the standard fundamental

domain of SL(2,Z), the proof of our Theorem then follows from this inequality.We abbreviate Θ(f) by Θ. Let, as before, f =

∏ki=1(νix1 − µix2). We know

that there exist δ > 0 and ti > 0 such that

f =√

Θδk/4

∏(tiνix1 − tiµix2)

and δ is the determinant PR−Q2 of the quadratic form

Px21 − 2Qx1x2 +Rx2

2 =k∑

i=1

t2i (νix1 − µix2)(νix1 − µix2).

Note thatP =

∑t2i |νi|2, R =

∑t2i |µi|2.

This form also equals P (x1− zx2)(x1− zx2). Hence, when we write z = x+ iy,

Q = xP, R = P |z|2, δ = P 2y2.

Choose bi, ci ∈ C such that√Pbi = νiti and

√Rci = −µiti. Then

∑|bi|2 =∑

|ci|2 = 1 and also

f =√

Θ1

yk/2

∏(bix1 + ci|z0|x2)(bix1 + ci|z0|x2).

Comparison of the r-th coefficients yields(k

r

)ar =

(|z0|r

yk/2

)(∑

#S=k−r

bScS′)√

Θ.

Here the summation is over all subsets S of 1, . . . , k of cardinality k− r, and S′

is the complement of S. Furthermore bS denotes the product of all bi, i ∈ S.We first use Schwarz’s inequality ∑

#S=k−r

bScS′

2

∑#S=k−r

|bS |2 ∑

#S=k−r

|cS′ |2 .

Finally use the generalised AM/GM inequality to obtain

∑#S=k−r

|bS |2 ≤(

k

k − r

)(1k

∑i

|bi|2)k−r

=(k

r

)1

kk−r

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F.Beukers, The diophantine equation Axp + Byq = Czr 26

and similarly ∑#S=k−r

|cS′ |2 ≤(k

r

)1kr.

Combining all inequalities yields the desired estimate for |ar|.qed

Using the estimate of Θ(f) for any Hermite reduced f ∈ C5(d) we obtain thefollowing consequence.

Corollary 9.6 Let f ∈ C5(d) and suppose f is Hermite reduced. Let a0, . . . , a12

be its coefficients. Then, for every i, j with i+ j ≤ 12 we have

|aiaj | ≤ 21255|d|2.

In particular, |ai| ≤ 1600√

5|d| for every i ≤ 6.

10 An algorithm to solve x2 + y3 = dz5

Let d be any non-zero integer. We have seen in the previous two sections thatall coprime solutions x, y, z to x2 + y3 = dz5 arise as specialisation to integersof a form f ∈ C5(d) and its Hessian and Jacobian covariant. To determinethe set C5(d) it suffices to determine the SL(2,Z)-orbits within C5(d). Moreparticularly, it suffices to determine the Hermite reduced forms in C5(d).Here is an algorithm to find the Hermite reduced forms with a0 6= 0.

1. Let B = 1600√

5|d|.

2. For all a0, a1, a2 ∈ Z with |ai| ≤ B and a0 6= 0 we do the following.

(a) Let Z = a0, Y = a0a2 − a21.

(b) Determine the at most two solutions a3 of X = ±√−Y 3 − dZ5 and

a20a3 − 3a0a1a2 + 2a3

1 = 2X

(c) Compute a4, . . . , a12 from the equations defining C5(d).

(d) If all a3, . . . , 7a6, . . . , a12 are integers and if they satisfy the boundsof Corollary 9.6 then we output the form [a0, . . . , a12].

When a0 we follow a similar procedure, but now we can assume a1 6= 0. Thevalues of a3, a4, . . . follow from the equations D4 = 0, D5 = 0, . . ..We have now a finite set F of forms in C5(d). We like to keep only the Hermitereduced ones. For that we determine the representing point z(f) ∈ H for eachf ∈ H. This can be a tedious computation, but we use the following observation.Every form f ∈ C5(d) is GL(2,R)-equivalent to x1x2(x10

1 − 11x51x

52 − x10

2 ). Thelatter form has four real roots, hence any form in C5(d) has four real roots.Let f1 be the factor of f consisting of the four real linear factors of f . Then,by standard arguments as explained in [CS], it turns out that the representingpoint of f is the same as that of f1. For the latter there are standard formulas.

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F.Beukers, The diophantine equation Axp + Byq = Czr 27

We delete from F the non-Hermite reduced forms. We are now left with a fullset of representatives of the SL(2,Z)-orbits in C5(d).In the final listing it saves space to look at GL(2,Z)-orbits in C5(d). Supposewe have a form f which, together with its covariants H(f), t(f)/2 represents aset S of solutions to x2 + y3 = dz5. Let g ∈ GL(2,Z) and det(g). Then, by thecovariant property we have H(f g) = H(f) and t(f g) = −t(f). So the formf g represents the set (−x, y, z)|(x, y, z) ∈ S of solutions.Of course we also delete those f from F that do not give rise to coprime solutions.

11 Appendix A: Parametrizing X2 + Y 3 ± Zr = 0

This section has been taken directly from Johnny Edwards’s paper [Ed]. It givescomplete parametrizations to X2 +Y 3±Zr = 0 for r = 3, 4, 5. In the tables welist the forms

f =k∑

i=0

(k

i

)aix

k−i1 xi

2

by the corresponding vector

[a0, a1, . . . , ak]

where k = 4, 6, 12 if r = 3, 4, 5 respectively. From this form we can compute thecovariant forms

H =1

k2(k − 1)2

∣∣∣∣ f11 f12f21 f22

∣∣∣∣ , g =1

2k(k − 2)

∣∣∣∣ f1 f2H1 H2

∣∣∣∣ .Here fij means ∂2f

∂xi∂xjetc. The forms then satisfy g2 +H3 ± fr = 0 and each

give infinitely many integer primitive solutions of the corresponding diophan-tine equation by specialisation of the polynomial variables. Moreover, solutionsets given by different parametrisations are disjoint, and their union is the fullsolution set. To keep the lists as short as possible, we identify the parametriza-tions identifying ±X. If the corresponding GL(2,Z) class of f breaks into twoSL(2,Z) classes these are really 2 distinct parametrizations.The case r = 3 was already done by Mordell in [Mo], Chapter 25 using a syzygyfrom invariant theory. The cases r = 4 were done by Zagier and quoted in [Beu],appendix A. The r = 5 case is new and presented in [Ed].

Complete Parametrization of X2 + Y 3 + Z3 = 0

A1 = [0, 1, 0, 0,−4]A2 = [−1, 0, 0, 2, 0]B1 = [−2,−1, 0,−1,−2]B2 = [−1, 1, 1, 1,−1]C1 = [−1, 0,−1, 0, 3]C2 = [1, 0,−1, 0,−3]

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F.Beukers, The diophantine equation Axp + Byq = Czr 28

In Mordell’s book [Mo] he further shortens the list by assuming that Z is odd.This means that A1, B1 can be omitted. However, Mordell gives 5 parametriza-tions: A2, B2, C1, C2 and f = [−1,−2,−4,−6, 0] According to [Ed] the 5thshould be superfluous. It turns out that f(x1 − 2x2, x2) is A2In [Beu], on page 78, parametrizations obtained by interchanging Y and Z areidentified.

Complete Parametrization of X2 + Y 3 ± Z4 = 0

These two equations were solved by Zagier and quoted in [Beu]. In [Co] thereis a complete solution according to classical lines and the lines followed byZagier. To keep the lists short we identify ±X and ±Z. This means everyparametrization in the list is shorthand for ±f(x1,±x2). The first ± is the ±Z.

The equation X2 + Y 3 + Z4 = 0:

f1 = [0, 1, 0, 0, 0,−12, 0]f2 = [0, 3, 0, 0, 0,−4, 0]f3 = [−1, 0, 1, 0, 3, 0,−27]f4 = [−3,−4,−1, 0, 1, 4, 3]

The equation X2 + Y 3 − Z4 = 0:

f1 = [0, 1, 0, 0, 0, 12, 0]f2 = [0, 3, 0, 0, 0, 4, 0]f3 = [−1, 0, 0, 2, 0, 0, 32]f4 = [−1, 0,−1, 0, 3, 0, 27]f5 = [−1, 1, 1, 1,−1, 5, 17]f6 = [−5,−1, 1, 3, 3, 3, 9]f7 = [−7,−1, 2, 4, 4, 4, 8]

Complete Parametrization of X2 + Y 3 + Z5 = 0

Beukers in [Beu] was able to produce parametrizations, though his method wasunable to produce a complete set. If we identify ±X, we have the followingcomplete set:

f1 = [0, 1, 0, 0, 0, 0,−144/7, 0, 0, 0, 0,−20736, 0]f2 = [−1, 0, 0,−2, 0, 0, 80/7, 0, 0, 640, 0, 0,−102400]

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F.Beukers, The diophantine equation Axp + Byq = Czr 29

f3 = [−1, 0,−1, 0, 3, 0, 45/7, 0, 135, 0,−2025, 0,−91125]f4 = [1, 0,−1, 0,−3, 0, 45/7, 0,−135, 0,−2025, 0, 91125]f5 = [−1, 1, 1, 1,−1, 5,−25/7,−35,−65,−215, 1025,−7975,−57025]f6 = [3, 1,−2, 0,−4,−4, 24/7, 16,−80,−48,−928,−2176, 27072]f7 = [−10, 1, 4, 7, 2, 5, 80/7,−5,−50,−215,−100,−625,−10150]f8 = [−19,−5,−8,−2, 8, 8, 80/7, 16, 64, 64,−256,−640,−5632]f9 = [−7,−22,−13,−6,−3,−6,−207/7,−54,−63,−54, 27, 1242, 4293]f10 = [−25, 0, 0,−10, 0, 0, 80/7, 0, 0, 128, 0, 0,−4096]f11 = [6,−31,−32,−24,−16,−8,−144/7,−64,−128,−192,−256, 256, 3072]f12 = [−64,−32,−32,−32,−16, 8, 248/7, 64, 124, 262, 374, 122,−2353]f13 = [−64,−64,−32,−16,−16,−32,−424/7,−76,−68,−28, 134, 859, 2207]f14 = [−25,−50,−25,−10,−5,−10,−235/7,−50,−49,−34, 31, 614, 1763]f15 = [55, 29,−7,−3,−9,−15,−81/7, 9,−9,−27,−135,−459, 567]f16 = [−81,−27,−27,−27,−9, 9, 171/7, 33, 63, 141, 149,−67,−1657]f17 = [−125, 0,−25, 0, 15, 0, 45/7, 0, 27, 0,−81, 0,−729]f18 = [125, 0,−25, 0,−15, 0, 45/7, 0,−27, 0,−81, 0, 729]f19 = [−162,−27, 0, 27, 18, 9, 108/7, 15, 6,−51,−88,−93,−710]f20 = [0, 81, 0, 0, 0, 0,−144/7, 0, 0, 0, 0,−256, 0]f21 = [−185,−12, 31, 44, 27, 20, 157/7, 12,−17,−76,−105,−148,−701]f22 = [100, 125, 50, 15, 0,−15,−270/7,−45,−36,−27,−54,−297,−648]f23 = [192, 32,−32, 0,−16,−8, 24/7, 8,−20,−6,−58,−68, 423]f24 = [−395,−153,−92,−26, 24, 40, 304/7, 48, 64, 64, 0,−128,−512]f25 = [−537,−205,−133,−123,−89,−41, 45/7, 41, 71, 123, 187, 205,−57]f26 = [359, 141,−1,−21,−33,−39,−207/7,−9,−9,−27,−81,−189,−81]f27 = [295,−17,−55,−25,−25,−5, 31/7,−5,−25,−25,−55,−17, 295]

The GL(2,Z) classes of the 27 forms split into 2 distinct SL(2,Z) classes, unlessf = f3, f4, f17, f18, f27. This means that the above list becomes 49 parametriza-tions if we do not identify ±X.

12 Appendix B: fourth transvectants

In this appendix, again reproduced from [Ed], we reproduce the equations sat-isfied by f of any form satisfying g2 + H3 + dfr = 0, where r, g,H are as inAppendix A. These equations are obtained by setting the fourth transvectantof f equal to zero and a further equation to specify scaling. The expressions Di

are the coefficients of the fourth transvectant τ4(f) =∑2k−8

i=0 Dixr−i1 xi

2. Notethat in all cases to any such form there corresponds a solution X,Y, Z of the

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F.Beukers, The diophantine equation Axp + Byq = Czr 30

equation X2 + Y 3 + dZr = 0 by evaluation f,H, g at (1, 0),

Z = a0

Y = a0a2 − a21

2X = a20a3 − 3a0a1a2 + 2a3

1

The tetrahedral case r = 3

0 = a0a4 − 4a1a3 + 3a22

−4d = a0a2a4 + 2a1a2a3 − a32 − a0a

23 − a2

1a4

The octahedral case r = 4

D0/1 : 0 = a4a0 − 4a3a1 + 3a22

D1/2 : 0 = a0a5 − 3a1a4 + 2a3a2

D2/1 : 0 = a0a6 − 9a2a4 + 8a23

D3/2 : 0 = a1a6 − 3a2a5 + 2a3a4

−72d = a0a6 − 6a1a5 + 15a2a4 − 10a23

The last equation is obtained from τ6(f) = 72d.

The icosahedral case r = 5

D0/1 : 0 = a0a4 − 4a1a3 + 3a22

D1/8 : 0 = a0a5 − 3a1a4 + 2a3a2

D2/4 : 0 = a0(7a6)− 12a1a5 − 15a2a4 + 20a23

D3/56 : 0 = a0a7 − 6a2a5 + 5a3a4

D4/14 : 0 = 5a0a8 + 12a1a7 − 6a2(7a6)− 20a3a5 + 45a24

D5/56 : 0 = a0a9 + 6a1a8 − 6a2a7 − 4a3(7a6) + 27a4a5

D6/28 : 0 = a0a10 + 12a1a9 + 12a2a8 − 76a3a7 − 3a4(7a6) + 27a4a5

D7/8 : 0 = a0a11 + 24a1a10 + 90a2a9 − 130a3a8 − 405a4a7 + 60a5(7a6)D8/1 : 0 = a0a12 + 60a1a11 + 534a2a10 + 380a3a9 − 3195a4a8

−720a5a7 + 60(7a6)2

D9/8 : 0 = a1a12 + 24a2a11 + 90a3a10 − 130a4a9 − 405a5a8 + 60(7a6)2

By elimination of a6, a7 from D2 = D3 = D4 = 0 we get

D∗4 : a3

0a8 = 12a4a3a1a0 + 18a4a22a0 − 24a2

3a2a0 + 4a5a3a20 − 9a2

4.

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F.Beukers, The diophantine equation Axp + Byq = Czr 31

From τ6(f) = 360df we get by comparison of the coefficients of x121 and x11

1 x2,

R0/1 : 360da0 = a0(7a6)− 42a1a5 + 105a2a4 − 70a23

R1/6 : 720da1 = 7a0a7 − 5a1(7a6) + 63a2a5 − 35a3a4

13 References

[BCDT ] C.Breuil, B.Conrad, F.Diamond, R.Taylor, On the modularity of ellipticcurves over Q: wild 3-adic exercises, J.Amer.Math.Soc. 14 (2001), 843-939.

[Bec ] S. Beckmann, On extensions of number fields obtained by specializingbranched coverings, J.reine angew. Math. 419 (1991), 27-53.

[Ben ] M.A.Bennett, The equation x2n + y2n = z5.

[Beu ] F.Beukers, The diophantine equation Axp + Byq = Czr, Duke Math.J.91(1998), 61-88.

[Br1 ] N.Bruin, The diophantine equations x2 ± y4 = ±z6 and x2 + y8 = z3,Compositio Math. 118 (1999), 305-321.

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